Problem: Let N be a positive odd number.

There are N coins, numbered 1, 2, \ldots, N. For each i (1 \leq i \leq N), when Coin i is tossed, it comes up heads with probability p_i and tails with probability 1 - p_i.

Taro has tossed all the N coins. Find the probability of having more heads than tails.

Constraints

* N is an odd number.
* 1 \leq N \leq 2999
* p_i is a real number and has two decimal places.
* 0 < p_i < 1

Input

Input is given from Standard Input in the following format:


N
p_1 p_2 \ldots p_N


Output

Print the probability of having more heads than tails. The output is considered correct when the absolute error is not greater than 10^{-9}.

Examples

Input

3
0.30 0.60 0.80


Output

0.612


Input

1
0.50


Output

0.5


Input

5
0.42 0.01 0.42 0.99 0.42


Output

0.3821815872

Solution (in JAVA):

Solution:
import java.util.Arrays;
import java.util.Scanner;

class Main{
	static int N;
	static double [] p;
	static double [][] memo;

	public static void main(String[] args) {
		Scanner sc = new Scanner(System.in);
		N = sc.nextInt();

		p = new double[N];

		memo = new double[N][(N / 2) + 1];
		for(int i = 0; i < N; i++) {
			p[i] = sc.nextDouble();
			Arrays.fill(memo[i], -1);
		}

		System.out.printf("%.10f\n",1 - dp(0,(N / 2)));
		sc.close();
	}

	static double dp(int i, int n) {
		if(i == N && n >= 0) {
			return 1;
		}
		if(n < 0) {
			return 0;
		}
		if(memo[i][n] != -1) {
			return memo[i][n];
		}
		memo[i][n] = (dp(i + 1, n) * (1 - p[i])) + (dp(i + 1, n - 1) * p[i]);
		return memo[i][n];
	}
}